poj3278——bfs

POJ 3278   对数轴进行一维bfs

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 52161		Accepted: 16355

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a pointK (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Input

Line 1: Two space-separated integers:  N and  K

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Input

5 17

Sample Output

4

题意：在数轴上求规定走法的最短路
思路：bfs
很基础的一个bfs，一开始由于边界没控制好RE了好多次，最后发现是处理边界的时候由于语句顺序不对导致vis数组越界了，最终还是AC了过去

//poj3278_bfs

#include<iostream>

#include<cstdlib>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<queue>



using namespace std;



const int maxn=100020;

int n,k;

bool vis[maxn];

struct node

{

    int pos,step;

};



int bfs()

{

    memset(vis,0,sizeof(vis));

    queue<node> q;

    q.push({n,0});

    vis[n]=1;

    while(!q.empty()){

        node now=q.front();

        q.pop();

        if(now.pos==k) return now.step;

        int next;

        //move to now.pos-1

        next=now.pos-1;

        if(next>=0&&next<maxn&&!vis[next]){ //控制边界时注意要将vis放在next<maxn之后，避免数组越界

            q.push({next,now.step+1});

            vis[next]=1;

        }

        //move to now.pos+1

        next=now.pos+1;

        if(next<maxn&&!vis[next]){

            q.push({next,now.step+1});

            vis[next]=1;

        }

        //move to now.pos*2

        next=now.pos*2;

        if(next<maxn&&!vis[next]&&next!=0){

            q.push({next,now.step+1});

            vis[next]=1;

        }

    }

    return false;

}



int main()

{

    while(cin>>n>>k){

        cout<<bfs()<<endl;

    }

    return 0;

}

poj3278_bfs